Decimation and the One-Dimensional Ising RG

The one-dimensional Ising model is the simplest interacting many-body system, and it famously refuses to order at any temperature above absolute zero. You can prove this several ways — a domain-wall entropy argument, the exact transfer matrix — but the renormalization group offers something the others do not: it shows you the machinery of forgetting. Sum a chosen half of the spins out of the partition function and what remains is another Ising chain, weaker-coupled, on a lattice twice as coarse. Repeat, and the coupling bleeds toward zero. The chain literally forgets its own alignment as you step back from it. Because the chain is exactly solvable, every one of those steps can be done in closed form. There is no truncation, no small parameter, no uncontrolled approximation — the recursion relation is exact. That makes decimation the cleanest laboratory in all of statistical mechanics for the ideas the renormalization group is built from: coarse-graining, rescaling, fixed points, and the flow between them. Everything that in higher dimensions requires clever schemes and $\epsilon$ -expansions is here reduced to a single line of algebra you can check by hand. The payoff is a constructive answer to why there is no phase transition. A critical point is a place where the correlation length is infinite, and the correlation length is the one quantity coarse-graining shrinks by the rescaling factor at every step. To sit still under the flow — to be critical — the chain must sit at a fixed point, and in one dimension the only fixed point with infinite correlation length is at $T=0$ . Criticality is banished to zero temperature not by fiat but by the arithmetic of squaring a number smaller than one.

Key takeaways

The 1D Ising chain can be renormalized exactly by decimation: sum over every other spin in the partition function, leaving an effective chain of half as many spins with a renormalized coupling.
The exact recursion is $K' = \tfrac12\ln\cosh 2K$ (equivalently $\tanh K' = \tanh^2 K$ ), where $K=\beta J$ , plus a known additive free-energy constant.
Iterating drives $K\to 0$ for any finite temperature: the only stable fixed point is the disordered one, so there is no phase transition at $T>0$ .
The unstable fixed point sits at $K=\infty$ ( $T=0$ ); the correlation length diverges only there, as $\xi\sim\tfrac12 e^{2K}$ — an essential singularity, so $\nu=\infty$ .
Summing the accumulated free-energy constants over the whole flow reconstructs the exact $f=\ln(2\cosh K)$ : the "bookkeeping" constants are the thermodynamics.
Key idea: summing over alternate spins is an exact RG step mapping an Ising chain at $K$ onto one at $K'$ with $\tanh K'=\tanh^2 K$ ; the macroscopic physics is read straight off the flow toward the trivial fixed points $K=0$ and $K=\infty$ .

The core insight

Summing over alternate spins is an exact RG step: it maps an Ising chain at coupling $K$ onto another at $K'$ with $\tanh K'=\tanh^2 K$ and a known free-energy shift, so the macroscopic physics is read off from the flow toward the trivial fixed points $K=0$ and $K=\infty$ .

Where this lives today

Exact decimation is the first RG calculation every graduate student does, precisely because it strips the method to its skeleton: coarse-grain, rescale, read the flow. But it is not merely pedagogical. Decimation becomes exact on hierarchical (diamond) lattices, where it produces genuine finite-temperature transitions with computable exponents — these solvable lattices are the testbeds where new RG ideas are first checked. The same bond-decimation logic, applied to the strongest bond or field at each step rather than to alternating sites, is the Ma–Dasgupta–Hu strong-disorder RG that governs random quantum spin chains and their infinite-randomness fixed points. And the block-decimation idea, with the classical "block spin" replaced by an optimally truncated quantum Hilbert space, is the ancestor of the density-matrix renormalization group (DMRG), today's workhorse for one-dimensional quantum many-body systems. The humble act of summing out every other spin turns out to seed an entire family of modern methods.

The physical picture

Picture the chain as alternating "kept" and "doomed" spins. Each doomed spin sits between two kept neighbours; summing it out of the partition function replaces the two bonds it bridged with a single effective bond directly between the survivors. That effective bond is weaker than either original — information about alignment leaks away when you forget the spin in the middle. Iterate the operation and the effective coupling bleeds toward zero: at large scales, the chain has forgotten its order. This is the same coarse-grain–and–rescale program set out in the renormalization group idea, made completely explicit. Decimation is just block-spin RG with block size $b=2$ and the crudest possible projection — keep every second site, discard the rest — but in one dimension that crude choice happens to be exact.

Figure

Decimation with rescaling factor b＝2: sum out the doomed spins, keep the survivors, close up the lattice

The correlation length is the one length the RG cannot manufacture. Measured in physical units it is what it is; but measured in lattice units it halves every step, because each step doubles the lattice spacing. The only way to have $\xi=\infty$ — a genuine critical point — is to sit exactly where the recursion has nothing left to do, a fixed point. In one dimension that demands $T=0$ . The cleanest way to see the whole phase diagram is to watch the variable $\tanh K\in(0,1)$ . Each decimation squares it. Squaring a positive number less than one drives it relentlessly to zero — unless it is exactly one. That single algebraic fact, $t\mapsto t^2$ , encodes everything: any finite temperature ( $\tanh K<1$ ) flows to disorder, and only the $T=0$ endpoint ( $\tanh K=1$ ) stands still.

Figure

The recursion tanh K′ ＝ tanh²K squares a number in (0,1): a cobweb flowing to the disordered fixed point

Finally, do not dismiss the additive free-energy constant that each step generates as bookkeeping clutter. The coupling flows to a trivial fixed point and stops carrying information — but the accumulated constants, summed over every RG step, reconstruct the exact free energy of the original chain. In one dimension you can watch the thermodynamics being assembled, term by term, out of what looks like a throwaway prefactor.

The core equations

We work with the zero-field Ising chain of $N$ spins and periodic boundaries. The dimensionless coupling $K=\beta J=J/k_BT$ absorbs the temperature, so "high temperature" means small $K$ and "low temperature" means large $K$ . Everything below is exact.

Z=\sum_{\{\sigma_i\}}\exp\!\Big(K\sum_i \sigma_i \sigma_{i+1}\Big),\qquad K=\beta J,\ \sigma_i=\pm1 \qquad (1)

where: $K$ — dimensionless nearest-neighbour coupling; $\sigma_i$ — Ising spin at site $i$ , taking $\pm1$ ; $Z$ — canonical partition function. Read it as a product of bond weights $e^{K\sigma_i\sigma_{i+1}}$ : each bond rewards alignment (both spins parallel) and penalizes anti-alignment. The decimation is a single spin sum. Take a doomed spin $\sigma_2$ sandwiched between survivors $\sigma_1$ and $\sigma_3$ . It appears in exactly two bonds, and nowhere else, so its sum is independent of everything happening elsewhere in the chain. Summing over its two values gives a hyperbolic cosine.

\sum_{\sigma_2=\pm1} e^{K \sigma_1 \sigma_2 + K \sigma_2 \sigma_3}=2\cosh\!\big(K(\sigma_1+\sigma_3)\big) \qquad (2)

where: $\sigma_2$ — the decimated spin between the survivors $\sigma_1,\sigma_3$ . The result depends on $\sigma_1,\sigma_3$ only through their sum $\sigma_1+\sigma_3\in\{-2,0,+2\}$ — which is the loophole that lets us rewrite it as a brand-new Ising bond. We demand that the result take the form of a renormalized two-spin Boltzmann weight, an Ising bond of new strength $K'$ times a spin-independent constant. Consistency for both distinct values of $\sigma_1\sigma_3$ fixes both unknowns.

2\cosh\!\big(K(\sigma_1+\sigma_3)\big)=A\,e^{K' \sigma_1 \sigma_3},\qquad A=e^{2g} \qquad (3)

where: $K'$ — renormalized coupling on the halved chain; $A$ — the spin-independent multiplicative constant, and $g$ the associated per-decimated-spin free-energy constant. Solving the two matching conditions gives the two central results of the whole article: the renormalized coupling and the free-energy constant. Both are exact.

\boxed{\;e^{2K'}=\cosh 2K\quad\Longleftrightarrow\quad K'=\tfrac12\ln\cosh 2K\;},\qquad A=2\sqrt{\cosh 2K} \qquad (4)

where: $K\mapsto K'$ — the exact RG map for rescaling factor $b=2$ ; $A=e^{2g}$ with $g=\tfrac12\ln\!\big(2\sqrt{\cosh 2K}\big)$ . Since $\cosh 2K\ge1$ , we always have $K'\ge0$ ; and since $\cosh 2K < e^{2K}$ , we always have $K'<K$ — the coupling can only weaken. The recursion has a famous compact form in the natural variable $\tanh K$ , which makes the flow transparent: decimation simply squares $\tanh K$ .

\tanh K' = \tanh^2 K \qquad (5)

where: the map $t\mapsto t^2$ with $t=\tanh K\in(0,1)$ . Because $0<t<1$ for any finite $T>0$ , iterating drives $t\to0$ , i.e. $K\to0$ — the disordered fixed point. This is the compact, equivalent restatement of $(4)$ ; the derivation below shows the two forms agree. The free-energy constants are where the thermodynamics hides. Each decimated spin contributes $g(K)$ to $\ln Z$ , and the free energy per spin obeys its own recursion tied to the halving of the spin count.

f(K)=g(K)+\tfrac12\,f(K'),\qquad g(K)=\tfrac12\ln 2+\tfrac14\ln\cosh 2K,\qquad f=\lim_{N\to\infty}\tfrac1N\ln Z \qquad (6)

where: $f$ — dimensionless free energy per spin; the factor $\tfrac12$ because the rescaled chain has half as many spins. Iterating this and summing the $g$ 's reconstructs the exact $f(K)=\ln(2\cosh K)$ . Finally, the correlation length. It contracts by the rescaling factor $b=2$ under one step when measured in lattice units, and this scaling plus the exact form pins down $\xi$ completely.

\xi^{-1}(K)=-\ln\tanh K=\ln\coth K,\qquad \xi(K')=\tfrac12\,\xi(K),\qquad \xi\xrightarrow{K\gg1}\tfrac12 e^{2K} \qquad (7)

where: $\xi$ — correlation length in units of the original lattice spacing. It is finite for every $T>0$ and diverges only as $T\to0$ , confirming criticality only at $T=0$ .

The core derivations

Derivation ASolving the matching conditions for the exact recursion

We want to turn the single-spin sum $(2)$ into a genuine RG map. The claim is that summing out one spin between two survivors reproduces an Ising bond of renormalized strength $K'$ times a constant $A$ , and that consistency across the two inequivalent survivor configurations fixes both. The strategy is to evaluate equation $(3)$ separately for aligned survivors ( $\sigma_1\sigma_3=+1$ ) and anti-aligned survivors ( $\sigma_1\sigma_3=-1$ ), producing two equations in the two unknowns $A,K'$ , then take their ratio and product. The one subtlety to watch: the left side depends on $\sigma_1+\sigma_3$ , which is $\pm2$ when aligned and $0$ when anti-aligned — a distinction the right side must reproduce through the sign of $K'\sigma_1\sigma_3$ .

Step 1. Evaluate $(3)$ for aligned survivors, $\sigma_1=\sigma_3$ , so $\sigma_1+\sigma_3=\pm2$ and $\sigma_1\sigma_3=+1$ :

2\cosh 2K = A\,e^{K'}.

Step 2. Evaluate $(3)$ for anti-aligned survivors, $\sigma_1=-\sigma_3$ , so $\sigma_1+\sigma_3=0$ (hence $\cosh 0=1$ ) and $\sigma_1\sigma_3=-1$ :

2 = A\,e^{-K'}.

Step 3. Divide Step 1 by Step 2 to eliminate $A$ :

e^{2K'}=\cosh 2K\quad\Longrightarrow\quad K'=\tfrac12\ln\cosh 2K.

Why this is the coupling and not the constant: the ratio cancels the spin-independent prefactor, isolating precisely the part of the weight that depends on $\sigma_1\sigma_3$ — the new bond.

Step 4. Multiply Step 1 by Step 2 to isolate $A$ :

A^2 = 4\cosh 2K\quad\Longrightarrow\quad A=2\sqrt{\cosh 2K},

and with $A=e^{2g}$ , $g=\tfrac12\ln\!\big(2\sqrt{\cosh 2K}\big)=\tfrac12\ln2+\tfrac14\ln\cosh 2K$ . Why $A>0$ matters: it is a legitimate Boltzmann prefactor, contributing to the analytic (non-singular) part of the free energy — never a sign or a phase.

Step 5. Confirm the compact form. Using $e^{2K'}=\cosh 2K$ ,

\tanh K' = \frac{e^{2K'}-1}{e^{2K'}+1}=\frac{\cosh 2K-1}{\cosh 2K+1}=\frac{2\sinh^2 K}{2\cosh^2 K}=\tanh^2 K,

where the last step uses $\cosh 2K-1=2\sinh^2K$ and $\cosh 2K+1=2\cosh^2K$ . Verified: `verify_decimation-1d-ising-rg.py` checks $(2)$ for all $(\sigma_1,\sigma_3)$ , solves the matching to $e^{2K'}=\cosh2K$ and $A=2\sqrt{\cosh2K}$ symbolically, and confirms $\tanh K'=\tanh^2 K$ identically (checks 1–3).

Go deeper · Derivation B · 1 minFixed points and the absence of a finite-temperature transition

Derivation BFixed points and the absence of a finite-temperature transition

Now we extract the physics: locate the fixed points of the recursion and classify their stability, and thereby prove constructively that the Ising chain has no phase transition at any $T>0$ . A fixed point $K^*$ satisfies $K^*=R(K^*)$ , which in the $t=\tanh K$ variable reads $t^*=(t^*)^2$ . We solve that, then linearize the map near each root to decide whether nearby couplings are drawn in (stable) or expelled (unstable). The lesson is that criticality requires a fixed point with infinite correlation length, and we will find one only at $T=0$ .

Step 1. Fixed-point condition in the $t=\tanh K$ variable:

t^* = (t^*)^2 \quad\Longrightarrow\quad t^*(1-t^*)=0,

so $t^*=0$ (i.e. $K^*=0$ , $T=\infty$ ) or $t^*=1$ (i.e. $K^*=\infty$ , $T=0$ ). There is no root at any intermediate $0<t^*<1$ .

Step 2. Stability of $K^*=0$ . Near small $K$ , $R(K)=\tfrac12\ln\cosh 2K\approx K^2$ , so

\left.\frac{dK'}{dK}\right|_{K=0}=\tanh(2\cdot0)=0<1.

A slope below one means the fixed point is attractive: any finite temperature is pulled toward $K=0$ , disordering the chain at large scales. Why the derivative is $\tanh 2K$ : differentiate $R(K)=\tfrac12\ln\cosh 2K$ to get $R'(K)=\tanh 2K$ , which runs from $0$ at $K=0$ up toward $1$ as $K\to\infty$ .

Step 3. Stability of $K^*=\infty$ . For large $K$ , $\cosh 2K\approx\tfrac12 e^{2K}$ , so

R(K)\approx \tfrac12\ln\!\big(\tfrac12 e^{2K}\big)=K-\tfrac12\ln2.

The slope approaches $1$ from below (marginal), and since $R(K)<K$ the flow drifts down toward smaller $K$ . So $K=\infty$ is the unstable (critical) fixed point.

Step 4. Conclusion. There is no fixed point at finite nonzero $K$ , hence no critical temperature $0<T_c<\infty$ . The chain's only critical point sits at $T=0$ , with the correlation length diverging there and nowhere else. Verified: `verify_decimation-1d-ising-rg.py` solves $t^*\in\{0,1\}$ , confirms $R'(0)=0$ (attractive) and $R(K)\to K-\tfrac12\ln2$ as $K\to\infty$ (marginal), and iterates $K_0=1.0$ to $<10^{-6}$ in 40 steps (checks 5–6).

Go deeper · Derivation C · 1 minCorrelation length and the essential singularity at T＝0

Derivation CCorrelation length and the essential singularity at T＝0

The RG scaling of $\xi$ combined with the recursion pins down the exact correlation length and reveals its unusual divergence. The goal is to derive $\xi^{-1}=-\ln\tanh K$ — the exact result also obtained from the transfer matrix — using only the RG contraction rule and the recursion, and then to extract the low-temperature asymptotics. The key physical input is that one decimation step doubles the lattice spacing, so a fixed physical length shrinks by a factor of two when re-expressed in the new lattice units: $\xi(K')=\tfrac12\xi(K)$ .

Step 1. Impose the RG contraction. Under $b=2$ ,

\xi(K')=\tfrac12\,\xi(K)\quad\Longleftrightarrow\quad \xi^{-1}(K')=2\,\xi^{-1}(K).

Step 2. Test the ansatz $\xi^{-1}(K)=-\ln\tanh K$ against the recursion $\tanh K'=\tanh^2K$ :

\xi^{-1}(K')=-\ln\tanh K'=-\ln\tanh^2 K=-2\ln\tanh K=2\,\xi^{-1}(K).

The ansatz satisfies the contraction rule identically — it is the exact correlation length. Why $-\ln\tanh K$ and not something else: the transfer-matrix eigenvalue ratio is $\lambda_-/\lambda_+=\tanh K$ , and correlations decay as $(\lambda_-/\lambda_+)^r=e^{-r/\xi}$ , giving $\xi^{-1}=-\ln\tanh K$ . The RG fixes the same object without diagonalizing anything.

Step 3. Low-temperature limit. For $K\gg1$ , $\tanh K\approx1-2e^{-2K}$ , so

-\ln\tanh K\approx 2e^{-2K}\quad\Longrightarrow\quad \xi\approx\tfrac12 e^{2K}=\tfrac12 e^{2J/k_BT}.

Step 4. The nature of the divergence. Because $\xi$ grows like $e^{2K}$ — exponentially in $1/T$ rather than as a power of $T$ — the standard form $\xi\sim|t|^{-\nu}$ would require $\nu=\infty$ . This is an essential singularity, the signature of a transition sitting exactly at the lower critical dimension. Verified: `verify_decimation-1d-ising-rg.py` confirms $\xi^{-1}(K')=2\xi^{-1}(K)$ (symbolic + numeric), $\xi\approx\tfrac12 e^{2K}$ at $K=6$ to $0.1\%$ , and $\ln\xi\approx 2K$ (slope $2.000$ ) certifying the essential singularity (checks 7–8, 13).

Limiting cases

The exact recursion lets us read off every regime analytically — no approximation is hiding anywhere.

Case	Result	Meaning
High temperature, $K\to0$	$\tanh K'\approx K^2\Rightarrow K'\approx K^2\to0$	Stable disordered fixed point; weak coupling becomes weaker and $\xi\to0$ . Generic high- $T$ behaviour.
Low temperature, $K\to\infty$	$K'\approx K-\tfrac12\ln2$	Near the $T=0$ fixed point the coupling decreases only slowly; that slow approach is exactly what makes $\xi$ exponentially large.
Correlation length	$\xi(K)\simeq\tfrac12 e^{2K}=\tfrac12 e^{2J/k_BT}$	Diverges only as $T\to0$ . Confirms no finite- $T$ transition; criticality sits at $T=0$ with $\nu=\infty$ .
Free energy	$f(K)=\ln(2\cosh K)$	Recovered exactly by summing the free-energy constants $g(K)$ along the flow — the constants carry the thermodynamics.

Common trapThe additive constant is not optional, and a field breaks the closure

Two traps. First, the additive free-energy constant $g(K)$ is not disposable bookkeeping: it accumulates the analytic part of the free energy, and dropping it means you never recover the correct thermodynamics — the coupling alone flows to a trivial fixed point and would tell you the free energy is trivial, which is false. Second, decimation in zero field closes on a single coupling $K$ , but the moment you turn on a magnetic field the recursion generates a renormalized field and a field–bond mixing: three inequivalent survivor configurations now require three parameters. The map still closes in 1D — each removed spin touches only two survivors — but in $d>1$ that same mechanism spawns further-neighbour and multi-spin couplings, which is precisely why real-space RG in higher dimensions must be approximate.

Worked examples

Example 1 — One decimation step reproduces the exact partition function of a short chain. We verify, with an actual number, that decimation is exact: a periodic 4-spin chain decimated down to a periodic 2-spin chain must give the identical $Z$ , up to the constant $A$ .

Step 1. Choose $K=0.6$ . Compute the renormalized coupling: $K'=\tfrac12\ln\cosh(1.2)=\tfrac12\ln(1.8107)=0.2967$ , and the constant $A=2\sqrt{\cosh 1.2}=2\sqrt{1.8107}=2.6912$ .

Step 2. Enumerate the periodic 4-spin chain directly. Summing $e^{0.6\,(\sigma_1\sigma_2+\sigma_2\sigma_3+\sigma_3\sigma_4+\sigma_4\sigma_1)}$ over all $2^4=16$ configurations gives $Z_4=34.228$ .

Step 3. Enumerate the decimated periodic 2-spin chain at $K'=0.2967$ : $Z_2=\sum e^{K'(\sigma_1\sigma_2+\sigma_2\sigma_1)}=2e^{2K'}+2e^{-2K'}=4.7256$ .

Step 4. Assemble. Decimating removes two spins, contributing $A^2$ : $A^2 Z_2 = (2.6912)^2\times 4.7256=7.2426\times4.7256=34.228$ . This equals $Z_4$ exactly (to all digits computed). The decimation lost nothing. $Z_4 = A^2 Z_2 = 34.228$ .

Example 2 — How many coarse-grainings to shrink the correlation length below one lattice spacing. At low temperature the chain is correlated over many sites; the RG tells us exactly how many blocking steps erase that.

Step 1. Take $K=2.0$ (well into the ordered-looking regime). The exact correlation length is $\xi=1/(-\ln\tanh 2.0)=1/(-\ln 0.96403)=1/0.036636=27.30$ lattice spacings.

Step 2. Each decimation halves $\xi$ in lattice units, so after $n$ steps $\xi_n=\xi/2^n$ . We need the smallest $n$ with $\xi/2^n<1$ .

Step 3. Solve $27.30/2^n<1\Rightarrow 2^n>27.30\Rightarrow n>\log_2 27.30=4.77$ , so $n=5$ . In closed form, $n=\lceil\log_2\xi\rceil=\lceil4.77\rceil=5$ .

Step 4. Sanity: after five steps $\xi_5=27.30/32=0.853<1$ , while after four $\xi_4=27.30/16=1.71>1$ . So five coarse-grainings reduce the chain from strongly correlated to effectively uncorrelated at the working scale — a concrete count of how far you must "step back" before the order dissolves. $n=5$ .

Origins · 1 minrecursion relation and how this was discovered

recursion relation

coarse-graining tool of the RG · The engine of every renormalization-group calculation is a recursion relation: a rule

K'=R(K)

that maps the couplings of a system at one length scale to those at a coarser scale. For the 1D Ising chain the recursion is exact,

\tanh K'=\tanh^2 K

, and the whole phase diagram follows from iterating it. The concept crystallized in the mid-1960s as Kadanoff turned the intuition of block spins into an explicit transformation, and reached full generality in Wilson's flows a few years later — a recursion relation is what makes the "group" in renormalization group.

History

1944

— Onsager's exact solution of the 2D Ising model shows a genuine finite-temperature transition, sharpening the puzzle of why 1D has none

1966

— Kadanoff introduces block-spin scaling for Ising models near T_c (Physics Physique Fizika 2, 263), the seed of real-space RG recursions

1971

— Wilson formulates the renormalization group as an explicit flow of couplings (two Physical Review B papers), making recursions like decimation instances of a general theory

1975

— Nelson and Fisher, and textbook treatments, establish exact 1D decimation with a field as the standard pedagogical RG

1982

— Wilson receives the Nobel Prize in Physics for the theory of critical phenomena in connection with phase transitions

Going further

References. Pathria & Beale, Statistical Mechanics, 3rd ed., ch. 14 (RG) · Kardar, Statistical Physics of Fields, ch. 4 · Reif, Fundamentals of Statistical and Thermal Physics · Sethna, Statistical Mechanics: Entropy, Order Parameters, and Complexity, ch. 12 · Landau & Lifshitz, Statistical Physics Part 1, §151 · Goldenfeld, Lectures on Phase Transitions and the Renormalization Group, ch. 9.

Provenance. All formulae are standard exact results for the 1D Ising model. A verifier should: (i) reproduce $e^{2K'}=\cosh2K$ and $\tanh K'=\tanh^2K$ from the single-spin sum; (ii) confirm the fixed points $K^*=0,\infty$ and their stability; (iii) check $\xi^{-1}=-\ln\tanh K$ against the transfer-matrix eigenvalues $\lambda_\pm=e^K\pm e^{-K}$ and the asymptotic $\xi\simeq\tfrac12 e^{2K}$ ; (iv) verify $f=\ln(2\cosh K)$ from the free-energy recursion; (v) check the field recursion reduces to zero-field as $h\to0$ . No fabricated constants; only textbook exact values are quoted.

Builds on: The Renormalization Group Idea: Coarse-Graining and Rescaling · Block-Spin (Real-Space) Renormalization Group · One-Dimensional Ising Model and Transfer Matrix Method · Fixed Points of the RG. Leads to: RG Flow in Coupling Space · Linearization About a Fixed Point and Eigenvalue Spectrum · Relevant, Irrelevant, and Marginal Operators · Critical Exponents from the RG.

Is decimation the same thing as block-spin RG?

It is the simplest special case. Block-spin RG groups $b^d$ spins per block and assigns each block a single coarse-grained spin by some rule (majority vote, weighted projection). Decimation with block size $b=2$ makes the crudest possible choice — keep every second spin, sum out the rest — which is not a "majority" of anything, just a selection. What makes 1D special is that this crude selection is exact: each removed spin couples only two survivors, so no approximation is needed. In higher dimensions the same selection would generate uncontrolled new couplings, and one must use genuine block averages plus truncation.

Why is there no finite-temperature transition even though the ground state is perfectly ordered?

At $T=0$ the chain is fully aligned — perfect order. But any $T>0$ admits domain walls, and in one dimension a single wall costs a fixed energy $2J$ while gaining an entropy $k_B\ln N$ from the $N$ places it can sit. For large $N$ the entropy term always wins, so walls proliferate and long-range order is destroyed at every positive temperature. The RG says the same thing in flow language: $\tanh K<1$ gets squared toward zero, so the coarse-grained chain always looks disordered. Order survives only at the single point $T=0$ , which is why the correlation length diverges there and nowhere else.

What exactly is a "fixed point" and why does criticality require one?

A fixed point is a coupling $K^*$ that the RG map leaves unchanged: $K^*=R(K^*)$ . Physically it is a scale-invariant theory — it looks the same at every magnification. The correlation length transforms as $\xi\to\xi/b$ under coarse-graining, so at a fixed point $\xi=\xi/b$ , forcing $\xi=0$ or $\xi=\infty$ . Criticality is the state of infinite correlation length, so a critical point must be a fixed point with $\xi=\infty$ . In 1D the only such point is $K^*=\infty$ ( $T=0$ ); the other fixed point, $K^*=0$ , is the $\xi=0$ trivial (fully disordered) one.

The correlation length shrinks under coarse-graining — doesn't that mean order is being destroyed by the RG itself?

No — the RG does not change the physics, only your ruler. The physical correlation length is fixed; when you double the lattice spacing, that same physical length simply counts as half as many lattice units. Nothing about the actual state changes. The shrinkage of $\xi$ in lattice units is a statement about the description, and it is exactly why a fixed point (where $\xi$ in lattice units is stationary) must have $\xi=\infty$ or $0$ . Away from a fixed point, iterating drives the system toward one of these, revealing which phase it is in.

Where does the free energy come from if the coupling just flows to a trivial fixed point?

This is the subtle and beautiful part. The coupling $K$ does flow to the trivial fixed point $K=0$ , where the free energy is just $\ln 2$ per spin (free spins). But every decimation step also emits an additive constant $g(K)$ . The recursion $f(K)=g(K)+\tfrac12 f(K')$ accumulates these constants down the flow. Summing them all — with the correct factor $\tfrac12$ at each level for the halving of spin count — reconstructs the exact free energy $f(K)=\ln(2\cosh K)$ . So the thermodynamics is not in the endpoint of the flow but in the trail of constants left along the way.

Why is the RG eigenvalue at the T＝0 fixed point associated with

\nu=\infty

Near $K=\infty$ the natural scaling variable is $u=e^{-2K}\to0$ , and the correlation length behaves as $\xi\sim u^{-1}=e^{2K}$ . If you insist on the power-law form $\xi\sim|t|^{-\nu}$ with $t$ measuring the distance to criticality, you find the divergence is exponential rather than power-law — an essential singularity — which corresponds to $\nu=\infty$ . This is characteristic of a system sitting exactly at its lower critical dimension, where a transition is only marginally absent. It contrasts sharply with the 2D Ising model, whose $\xi\sim|t|^{-1}$ gives the finite $\nu=1$ .

Does decimation generalize to give real transitions in higher dimensions?

Not directly on ordinary lattices — there, exact decimation spawns infinitely many couplings and must be truncated, which is uncontrolled. But on hierarchical (diamond) lattices, decimation stays exact and does produce genuine finite-temperature fixed points with nontrivial, computable critical exponents. These serve as solvable caricatures of real lattices, and the closely related Migdal–Kadanoff bond-moving scheme uses the same spirit to give surprisingly good approximate exponents in $d\ge2$ . So the 1D calculation is the conceptual seed of an entire family of tractable higher-dimensional RG models.